sd MANUAL
, 2-1
CHAPTER 2 sd
The correspondence between the problem set in this first edition versus the proble
sd sd sd sd sd sd sd sd sd sd sd sd
m set in the 5'th edition text. Problems that are new are marked new and those that a
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
re only slightly altered are marked as modified (mod).
sd sd sd sd sd sd sd sd
New Old New Old New Old
1 1 21 new 41 new
2 2 22 18 42 32
3 new 23 19 43 34
4 3 24 15 44 35
5 5 25 new
6 6 26 20
7 7 27 21
8 new 28 new
9 9 29 new
10 10 30 23
11 11 31 24
12 12 32 new
13 new 33 new
14 new 34 26
15 new 35 27
16 13 36 new
17 new 37 28mod
18 14 38 29
19 new 39 30
20 16 40 31
,2-2
2.1 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9
sd sd sd sd sd sd sd sd sd sd sd sd sd
.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this gr
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
avitational field ? How much mass can a force of 1 N support ?
sd sd sd sd sd sd sd sd sd sd sd sd sd
Solution:
ma = 0 = F = F - mg sd sd sd sd sd sd sd sd sd F
F = mg = 2 9.80665 = 19.613 N
s d sd sd sd sd sd sd sd sd
g
F = mg sd => sd
m
m = F/g = .80665 = 0.102 kg
sd sd sd sd sd sd sd sd sd
2.2 A model car rolls down an incline with a slope so the gravitational “pull” in the di
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
rection of motion is one third of the standard gravitational force (see Problem 2.1
sd sd sd sd sd sd sd sd sd sd sd sd sd
). If the car has a mass of 0.45 kg find the acceleration.
sd sd sd sd sd sd sd s d sd sd sd sd
Solution:
ma = F = mg / 3 sd s d sd sd sd sd sd sd
a = mg / 3m = g/3
sd sd sd sd sd sd
= 9. = 3.27 m/s2
sd sd sd sd sd sd
g
This acceleration does not depend on the mass of the model car.
sd sd sd sd sd sd sd sd sd sd sd
2.3 A force of 125 N is applied to a mass of 12 kg in addition to the standard gravita
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
tion. If the direction of the force is vertical up find the acceleration of the mass.
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
Solution: x
Fup = ma = F – mg s d
sd sd sd sd sd
F
a = ( F – mg ) / m = F / m – g
sd sd sd sd sd sd sd sd sd sd sd sd sd sd
= 125/12 – 9.807 sd s d sd
g
m
= 0.61 ms-2 sd sd
2.4 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 se
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
conds. If the total car and driver mass is 1075 kg find the necessary force.
sd sd sd sd sd sd sd sd sd sd sd sd sd sd
, 2-3
Solution:
Acceleration is the time rate of change of velocity. sd sd sd sd sd sd sd sd
dV 60 1000
= 3.33 m/s2
sd sd
a= =
3600 5
sd sd sd
dt sd sd
ma = F ;
sd sd sd sd
Fnet = ma = 1075 3.333 = 3583 N
sd
sd sd sd sd sd sd sd sd
2.5 A 1200-
sd
kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed o
sd sd sd sd sd sd sd sd sd sd sd sd s d sd
sd
sd sd sd sd
f 75 km/h. What are the force and total time required?
sd sd sd sd sd sd sd sd sd sd
Solution:
dV => (75 20) 100 = 3.82 sec
V 0=
sd sd sd sd
a= = s d sd
t =
sd sd s d s d
sd s d
V
dt t a 3600 5 sd sd
F = ma = 1200 4 = 4800 N
sd sd sd sd sd sd sd sd sd
2.6 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What f
sd sd sd sd sd sd sd sd sd sd sd
sd
sd sd sd sd sd sd
orce is needed and what is the final velocity?
sd sd sd sd sd sd sd sd
Solution:
Constant acceleration can be integrated to get velocity.
sd sd sd sd sd sd sd
dV
=> dV = a dt => V = a t
s d s d
a=
sd
dt
s d sd sd sd sd sd sd
ds ds
V = a t = 3 10 = 30 m/s
sd sd sd => V = 30 m/ssd sd sd sd sd sd sd sd sd sd
F
F = ma = 950 3 = 2850
sd
sd sd sd sd sd sd sd sd sd
N
, 2-1
CHAPTER 2 sd
The correspondence between the problem set in this first edition versus the proble
sd sd sd sd sd sd sd sd sd sd sd sd
m set in the 5'th edition text. Problems that are new are marked new and those that a
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
re only slightly altered are marked as modified (mod).
sd sd sd sd sd sd sd sd
New Old New Old New Old
1 1 21 new 41 new
2 2 22 18 42 32
3 new 23 19 43 34
4 3 24 15 44 35
5 5 25 new
6 6 26 20
7 7 27 21
8 new 28 new
9 9 29 new
10 10 30 23
11 11 31 24
12 12 32 new
13 new 33 new
14 new 34 26
15 new 35 27
16 13 36 new
17 new 37 28mod
18 14 38 29
19 new 39 30
20 16 40 31
,2-2
2.1 The “standard” acceleration (at sea level and 45° latitude) due to gravity is 9
sd sd sd sd sd sd sd sd sd sd sd sd sd
.80665 m/s2. What is the force needed to hold a mass of 2 kg at rest in this gr
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
avitational field ? How much mass can a force of 1 N support ?
sd sd sd sd sd sd sd sd sd sd sd sd sd
Solution:
ma = 0 = F = F - mg sd sd sd sd sd sd sd sd sd F
F = mg = 2 9.80665 = 19.613 N
s d sd sd sd sd sd sd sd sd
g
F = mg sd => sd
m
m = F/g = .80665 = 0.102 kg
sd sd sd sd sd sd sd sd sd
2.2 A model car rolls down an incline with a slope so the gravitational “pull” in the di
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
rection of motion is one third of the standard gravitational force (see Problem 2.1
sd sd sd sd sd sd sd sd sd sd sd sd sd
). If the car has a mass of 0.45 kg find the acceleration.
sd sd sd sd sd sd sd s d sd sd sd sd
Solution:
ma = F = mg / 3 sd s d sd sd sd sd sd sd
a = mg / 3m = g/3
sd sd sd sd sd sd
= 9. = 3.27 m/s2
sd sd sd sd sd sd
g
This acceleration does not depend on the mass of the model car.
sd sd sd sd sd sd sd sd sd sd sd
2.3 A force of 125 N is applied to a mass of 12 kg in addition to the standard gravita
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
tion. If the direction of the force is vertical up find the acceleration of the mass.
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
Solution: x
Fup = ma = F – mg s d
sd sd sd sd sd
F
a = ( F – mg ) / m = F / m – g
sd sd sd sd sd sd sd sd sd sd sd sd sd sd
= 125/12 – 9.807 sd s d sd
g
m
= 0.61 ms-2 sd sd
2.4 A car drives at 60 km/h and is brought to a full stop with constant deceleration in 5 se
sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd sd
conds. If the total car and driver mass is 1075 kg find the necessary force.
sd sd sd sd sd sd sd sd sd sd sd sd sd sd
, 2-3
Solution:
Acceleration is the time rate of change of velocity. sd sd sd sd sd sd sd sd
dV 60 1000
= 3.33 m/s2
sd sd
a= =
3600 5
sd sd sd
dt sd sd
ma = F ;
sd sd sd sd
Fnet = ma = 1075 3.333 = 3583 N
sd
sd sd sd sd sd sd sd sd
2.5 A 1200-
sd
kg car moving at 20 km/h is accelerated at a constant rate of 4 m/s2 up to a speed o
sd sd sd sd sd sd sd sd sd sd sd sd s d sd
sd
sd sd sd sd
f 75 km/h. What are the force and total time required?
sd sd sd sd sd sd sd sd sd sd
Solution:
dV => (75 20) 100 = 3.82 sec
V 0=
sd sd sd sd
a= = s d sd
t =
sd sd s d s d
sd s d
V
dt t a 3600 5 sd sd
F = ma = 1200 4 = 4800 N
sd sd sd sd sd sd sd sd sd
2.6 A steel plate of 950 kg accelerates from rest with 3 m/s2 for a period of 10s. What f
sd sd sd sd sd sd sd sd sd sd sd
sd
sd sd sd sd sd sd
orce is needed and what is the final velocity?
sd sd sd sd sd sd sd sd
Solution:
Constant acceleration can be integrated to get velocity.
sd sd sd sd sd sd sd
dV
=> dV = a dt => V = a t
s d s d
a=
sd
dt
s d sd sd sd sd sd sd
ds ds
V = a t = 3 10 = 30 m/s
sd sd sd => V = 30 m/ssd sd sd sd sd sd sd sd sd sd
F
F = ma = 950 3 = 2850
sd
sd sd sd sd sd sd sd sd sd
N
